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  • The solution of the differential equation frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=frac{1}{ 1+x^{2} ^{2}} is (a) y 1+x^{2} =C+\tan ^{-1} x

The solution of the differential equation \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}} is

(a) y\left(1+x^{2}\right)=C+\tan ^{-1} x
(b) \frac{y}{1+x^{2}}=C+\tan ^{-1} x
(c) y \log \left(1+x^{2}\right)=C+\tan ^{-1} x
(d) y\left(1+x^{2}\right)=C+\sin ^{-1} x

Answers (1)

Ans: - The answer is the option (a) y\left(1+x^{2}\right)=C+\tan ^{-1} x

Explanation: -

\frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}} is

Here, P=\frac{2 x}{1+x^{2}}$ and $Q=\frac{1}{\left(1+x^{2}\right)^{2}}
\\ \therefore$ I.F. $=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$ \\$\therefore \text{ The general solution is}
$$ y\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \frac{1}{\left(1+x^{2}\right)^{2}}+C
\\\Rightarrow \quad y\left(1+x^{2}\right)=\int \frac{1}{1+x^{2}} d x+C$ \\$\Rightarrow$ $y\left(1+x^{2}\right)=\tan ^{-1} x+C

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